I'll start with the first problem... has solutions only if b and c are parallel, that is, if for some constant . So, we set . We can decompose x into a piece that's parallel to a and a piece that's perpendicular to a, i.e., with and for some constant .

Since the dot product is linear, this gives . Thus, we see that u can be any vector we like as long as it's perpendicular to a, and we must have . The set of solutions, then, is for all u orthogonal to a.

Second part is a little trickier, but doable. Think of this as an equation of the form , where L is a linear operator. In this case we can write this operator . Here I is the identity, and is the conjugate transpose of a, so applying it to a vector gives .

So, the equation we want to solve is . In order to solve it, we need to find an inverse of the operator L, because then the solution is just .

Now I can't remember if there's an easy way to see this, but there happens to be an inverse of the form for some vector u. Suppose this is true. Then we want . Now expand this out: . All that's left is to solve for u, and apply the inverse operator you get to b to get your final result.